Show that two lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,where $b_{1}, b_{2} \neq 0$,are perpendicular if $a_{1}a_{2} + b_{1}b_{2} = 0$.

The given lines can be written in slope-intercept form $y = mx + c$ as:
$y = -\frac{a_{1}}{b_{1}}x - \frac{c_{1}}{b_{1}}$ ..... $(1)$
$y = -\frac{a_{2}}{b_{2}}x - \frac{c_{2}}{b_{2}}$ ..... $(2)$
The slopes of lines $(1)$ and $(2)$ are $m_{1} = -\frac{a_{1}}{b_{1}}$ and $m_{2} = -\frac{a_{2}}{b_{2}}$ respectively.
Two lines are perpendicular if the product of their slopes is $-1$,i.e.,$m_{1} \cdot m_{2} = -1$.
Substituting the values of $m_{1}$ and $m_{2}$:
$(-\frac{a_{1}}{b_{1}}) \cdot (-\frac{a_{2}}{b_{2}}) = -1$
$\frac{a_{1}a_{2}}{b_{1}b_{2}} = -1$
$a_{1}a_{2} = -b_{1}b_{2}$
$a_{1}a_{2} + b_{1}b_{2} = 0$
Thus,the lines are perpendicular if $a_{1}a_{2} + b_{1}b_{2} = 0$.